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# If x not equal to (-y), then is (x-y)/ (x + y) > 1

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If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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03 Mar 2010, 06:22
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If x not equal to (-y), then is (x-y)/ (x + y) > 1

(1) x > 0
(2) y < 0
[Reveal] Spoiler: OA

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03 Mar 2010, 10:11
This is my first post to GMATCLUB & this is how i do it.

is (x-y)/ (x + y) > 1?
==>is x-y>x+y
==>boils down to ..is y<0?
and does not depend on x at all

so ans is (B)

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03 Mar 2010, 11:42
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sudip135 wrote:
If x not equal to (-y), then is (x-y)/ (x + y) > 1

1) x > 0
2) y < 0

given x is not equal to -y

stmnt1) x>0

let x = 100 > 0 and y = -50

then (100 - (-50))/ (100 + (-50)) = 3 which is greater than 1

let x = 1/2 > 0 and y = 10

then ((1/2) - 10)/ ((1/2) + 10) = -19/21 which is less than 1. Hence insuff

stmnt2) y<0

let y = -100 < 0 and x= 50
150/-50 = = -3 <1

let y = -500 < 0 and x = 1000
1500/500 = 3>1. hence insuff

even together they do not suffice. so will go with E

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03 Mar 2010, 13:06
anujairaj770 wrote:
This is my first post to GMATCLUB & this is how i do it.

is (x-y)/ (x + y) > 1?
==>is x-y>x+y
==>boils down to ..is y<0?
and does not depend on x at all

so ans is (B)

U can do this as you don't know whether x+y is a positive value or a negative value. Hence for unknown variables this is wrong!
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03 Mar 2010, 13:07
Would go with E... as by plugging values, both statements are also not sufficient.
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03 Mar 2010, 13:19
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$$\frac{x-y}{x+y}>1$$ ---> $$\frac{x+y-2y}{x+y}>1$$ ---> $$1 - \frac{2y}{x+y}>1$$ ---->

$$\frac{y}{x+y}<0$$

Even if both statements are right, x+y could be either positive or negative. So, E
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03 Mar 2010, 14:39

But i don't understand how this assumption is incorrect?

(x-y)/ (x + y) > 1
==> x-y>x+y

I was able to conclude to E only after substituting numbers

There must be a faster way!!

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03 Mar 2010, 15:20
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anujairaj770 wrote:

But i don't understand how this assumption is incorrect?

(x-y)/ (x + y) > 1
==> x-y>x+y

I was able to conclude to E only after substituting numbers

There must be a faster way!!

Given: $$\frac{x-y}{x+y}>1$$. When you are then writing $$x-y>x+y$$, you are actually multiplying both sides of inequality by $$x+y$$: never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if $$x+y>0$$ you should write $$x-y>x+y$$ BUT if $$x+y<0$$, you should write $$x-y<x+y$$ (flip the sign when multiplying by negative expression).

Given inequality can be simplified as follows: $$\frac{x-y}{x+y}>1$$ --> $$0>1-\frac{x-y}{x+y}<0$$ --> $$0>\frac{x+y-x+y}{x+y}$$ --> $$0>\frac{2y}{x+y}$$ --> we can drop 2 and finally we'll get: $$0>\frac{y}{x+y}$$.

Now, numerator is negative ($$y<0$$), but we don't know about the denominator, as $$x>0$$ and $$y<0$$ can not help us to determine the sign of $$x+y$$. So the answer is E.

Hope it helps.
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04 Mar 2010, 06:41
Thanks a bunch for the explanation!

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04 Mar 2010, 15:15
Given that (x-y)/(x+y) > 1,
we know that

if (x+y) > 0, x-y > x+y => -y > y => y<0
x+y > 0 => x > -y

if (x+y) < 0, x-y < x+y => -y < y => y>0
x+y < 0 => x < -y

Only under two conditions will satisfy (x-y)/(x+y) > 1:
1. y <0 & x > -y (yes, x > -y => x > 0, but x > 0 does not mean x is larger than -y )
OR
2. y >0 & x < -y

thus, E should be the choice

Hope my explanation is clear and correct. let me know it is wrong or confusing please.

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05 Apr 2012, 09:42
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RaviChandra wrote:
If x is not equal to –y, is (x – y) / (x + y) > 1?
(1) x > 0 (2) y < 0

Working with 1 on the right hand side is hard. It is better in case we have 0 on the right hand side.

Is (x – y) / (x + y) - 1 > 0 ?
Is -2y/(x+y) > 0 ?

For -2y/(x+y) to be positive, either both (-2y) and (x+y) should be positive or both should be negative.

Both (-2y) and (x+y) positive
y should be negative and x should be positive with greater absolute value than that of y (so that x+y is positive)
OR
Both (-2y) and (x+y) negative
y should be positive and x should be negative with greater absolute value than that of y (so that x+y is negative)

Both statements together tell us that x is positive and y is negative but they still do not tell us whether absolute value of x is greater than that of y. Hence, both statements together are not sufficient.

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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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19 Sep 2013, 22:25
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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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28 Oct 2014, 09:53
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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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06 Mar 2016, 20:44
Hi Karishma,

Here is how I solved the problem:

$$\frac{(x-y)}{(x+y)}>1$$

$$x-y>x+y>0$$
or
$$x-y<x+y<0$$

For case 1, solve $$x-y>x+y$$ we get $$y<0$$. Then solve $$x+y>0$$ we get $$x>(-y)$$. Together, we have $$x>0>y$$.

For case 2, solve $$x-y<x+y$$ we get $$y>0$$. Then solve $$x+y<0$$ we get $$x<(-y)$$. Together, we have $$x<0<y$$.

The two cases imply that x and y have opposite signs. Since statement (1) and statement (2) prove that x and y have oposite signs, hence sufficient.

After reading the solutions of Bunuel and you multiple times, I still cannot understand where I went wrong. Please point out the mistake.
Thank you very much!

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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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06 Mar 2016, 21:10
truongynhi wrote:
Hi Karishma,

Here is how I solved the problem:

$$\frac{(x-y)}{(x+y)}>1$$

$$x-y>x+y>0$$
or
$$x-y<x+y<0$$

For case 1, solve $$x-y>x+y$$ we get $$y<0$$. Then solve $$x+y>0$$ we get $$x>(-y)$$. Together, we have $$x>0>y$$.

For case 2, solve $$x-y<x+y$$ we get $$y>0$$. Then solve $$x+y<0$$ we get $$x<(-y)$$. Together, we have $$x<0<y$$.

The two cases imply that x and y have opposite signs. Since statement (1) and statement (2) prove that x and y have oposite signs, hence sufficient.

After reading the solutions of Bunuel and you multiple times, I still cannot understand where I went wrong. Please point out the mistake.
Thank you very much!

Hi,
after you have find some range, then substitute values to see if it fits in..
you have found out that x and y are of opposite sign..
say x is -ive, then y is positive..

1) let x= -3 and y=5..
$$\frac{(x-y)}{(x+y)}>1$$...
$$\frac{(-3-5}{(-3+5)}>1$$...
$$\frac{-8}{2}>1$$... ans NO

2) let x= -3 and y=2..
$$\frac{(x-y)}{(x+y)}>1$$..
$$\frac{(-3-2}{(-3+2)}>1$$..
$$\frac{-5}{-1}>1$$... ans YES

Insuff even when combined
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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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06 Jun 2016, 09:12
i solved by plugging in some numbers and managed to get the answer. Is that not a good approach?

clearly each statement alone is not sufficient.

Take 1 + 2 together

x >0 ; y < 0 ; (x-y) / (x+y) >1
x= 2 ; y = - 1; yes
x= 1/2; y = - 1; No

therefore E

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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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09 Nov 2016, 23:35
jusjmkol740 wrote:
If x not equal to (-y), then is (x-y)/ (x + y) > 1

(1) x > 0
(2) y < 0

Please find the solution as attached
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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1 [#permalink]

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03 Jan 2018, 12:40
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Re: If x not equal to (-y), then is (x-y)/ (x + y) > 1   [#permalink] 03 Jan 2018, 12:40
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